\[h(\mathbf{x})=\mathbf{x}-\begin{pmatrix}\phantom{-} 1\\-21\end{pmatrix},\quad
	g(\mathbf{x})=\begin{pmatrix}
		98 & -1 \\1&\phantom{-} 98
	\end{pmatrix}\mathbf{x},\quad
	h^{-1}(\mathbf{x})=\mathbf{x}+\begin{pmatrix}\phantom{-}1\\-21\end{pmatrix}.\solnmarksplus{2}{1 for formulas for \(h\) and \(h^{-1}\),\\ 1 for formula for \(g\)\\ (can be implicit in working)}\]

Hence
\begin{align*}k(\mathbf{x}) & =(h^{-1}\circ g \circ h)(\mathbf{x})                                                                                                                          \\&=h^{-1}(g(h(\mathbf{x})))\\
              %line3
                            & =h^{-1}\left( g \left(\mathbf{x}-\begin{pmatrix}\phantom{-} 1\\-21\end{pmatrix}\right)\right)                                                                 \\
              %line4
                            & =h^{-1}\left(\begin{pmatrix}
	                                           98 & -1 \\1&\phantom{-} 98
                                           \end{pmatrix}\left(\mathbf{x}-\begin{pmatrix} \phantom{-} 1\\-21\end{pmatrix}\right)\right)\solnmarksplus{1}{substituting formulas in correctly} \\
              %line5
                            & =h^{-1}\left(\begin{pmatrix}
	                                           98 & -1 \\1&\phantom{-} 98
                                           \end{pmatrix}\mathbf{x}-\begin{pmatrix}
	                                                                   98 & -1 \\1& \phantom{-} 98
                                                                   \end{pmatrix}\begin{pmatrix} \phantom{-} 1\\-21\end{pmatrix}\right)                                                      \\
              %line6
                            & =h^{-1}\left(\begin{pmatrix}
	                                           98 & -1 \\1& \phantom{-} 98
                                           \end{pmatrix}\mathbf{x}-\begin{pmatrix}3\\1\end{pmatrix}\right)\solnmarksplus{1}{product of matrix and vector}                                   \\
              %line7
\end{align*}
